Coefficient of $x$ in a finite expansion

Question

Question:

Find coefficient of $x$ in $$(1+x+2x^2+3x^2+...+nx^n)^2$$

My attempt:

I know about the expansion involving $(1+x)^{-1}$. It cleanly is visible in the above question but the qusestion is a finite expansion while $(1+x)^{-1}$ is infinite and thus unsusable.

I thought of using $(a+b+c+d+...)^2=(a^2+b^2+c^2+...)+2(\text{sum of pairwise products})$ but it really is a long and cumbersome method for an objective question.

Is there another binomial expansion - that I am missing - that could have helped me in this objective question? Please give some hints.

Answer 1

Imagine multiplying out the brackets, $(1 + x + x^2 + ...)(1 + x + x^2 + ...)$; we would do this by taking each term in the left bracket and multiplying it by each term in the right bracket in turn.

Notice that, the only times that this procedure produces an '$x$' term is when multiplying a '1' from one of the brackets, and an $x$ from the other - any other combination will give a higher power of $x$.

Thus, when the expression is expanded, the only '$x$' terms we get are $1\cdot x$ and $x\cdot 1$ (from multiplying the '1' in the left bracket by the $x$ in the right bracket and vice-versa).

Hence, the coefficient of $x$ is 1 + 1 = 2.

Note: As you can see in the first line, I have not used the fact that the expression being squared ends at $nx^n$. In fact, even if it were an infinite series, then the above argument would still hold formally (i.e. ignoring convergence issues). This is essentially because any higher powers of $x$ in the expression being squared cannot possibly contribute to the 'number of $x$s' obtained when the expression is squared.

Hence, if you really wanted, the binomial theorem as you mentioned could be used, though it is not really necessary (and you would have to work around some technicalities such as for which values the result would hold due to convergence issues).

Answer 2

Let $g(x)=(1+x+2x^2+3x^2+\cdots+nx^n)^2$. The coefficient of $x$ is $g(x)$ is $g'(0)$.

Now, $g(x)=(1+xf'(x))^2$, where $f(x)=x+x^2+x^3+\cdots+x^n$.

Therefore, $g'(x)=2(1+xf'(x))(xf''(x)+f'(x))$ and so $g'(0)=2f'(0)=2$.

The combinatorial way is to notice that the only way to get $x$ in $(1+x+2x^2+3x^2+\cdots+nx^n)(1+x+2x^2+3x^2+\cdots+nx^n)$ is $1\cdot x$ and $x \cdot 1$, which gives $2x$.

Answer 3

Hint: $(a_1+a_2+...+a_n)^2=a_1^2+a_2^2+...+a_n^2+2a_1a_2+2a_1a_3+...+2a_1a_n+2a_2a_3+...$

The only way to get $x$ is by $2a_1a_2=2\cdot 1 \cdot x$. So the coefficient is $2$.

To shorten this down you could write $f(x)=2x^2+...+nx^n$ $(1+x+f(x))^2=1^2+x^2+f(x)^2+2x+2f(x)+2xf(x)$. It is clear that $f(x)$ is of degree 2 hence the coefficient of $x$ is $2$.

Answer 4

By checking the first cases we can conjecture that the coefficient is always $2$. Proving by induction :

Base case $n=1$ trivial.

Induction step : $$ (1+x + x^2 + ... + nx^n + (n+1)x^{n+1})^2 = \left( \overbrace{(1+x + x^2 + ... + nx^n )}^{A} + \underbrace{(n+1)x^{n+1}}_{B} \right)^2 \\ = A^2 + 2AB + B^2 $$ Clearly there is no $x$ in $B^2$ and also the degree of $2AB$ is at least $n+1$ so there is no $x$ either in it. Therefore the only place where $x$ can be is in $A$ which by inductive hypothesis has coefficient $2$ therefore $x$ has coefficient $2$ in $A^2 + 2AB + B^2$.

Answer 5

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an expression.

We can write \begin{align*} [x](1+x+2x^2+3x^3+\cdots+nx^n)^2&=[x](1+x)^2\tag{1}\\ &=[x](1+2x+x^2)\tag{2}\\ &=2 \end{align*}

Comment:

• In (1) we restrict the right-hand side to summands up to $x^1$ since higher powers do not contribute to the coefficient of $x$.

• In (2) we multiply out and select the coefficient of $x$.